We are given a sample of size $n$ ($X_i$), the ordered statistics are as follows:
$Y_1\leq Y_2 \leq Y_3$. . . . .$Y_{n-1}\leq Y_n$
I am trying to establish the the probability density function of $\alpha$th ordered statistic i.e $f_{Y_{\alpha}}(y)$.
The solution goes as follows:
$f_{Y_{\alpha}}(y)=\lim_{\Delta y->0}\dfrac{F_{Y_{\alpha}}(y+\Delta y)-F_{Y_{\alpha}}(y)}{\Delta y}$
= $\lim_{\Delta y->0}\dfrac{P(y<X_i\leq y+\Delta y)}{\Delta y}$
= $lim_{\Delta y->0}\dfrac{P((\alpha -1)\:of\:X_i \leq y;one\:X_i \: in\: (y,y+\Delta];(n-\alpha)\:of\:X_i > y+\Delta y)}{\Delta y}$
Now, let $N$ be the number of $X_i 's$ such that $X_i\leq y$, thus $N$~$Bin(n,F_{X_i}(y))$.
Also, let $N^{'}$ be the number of $X_i 's$ such that $X_i> y+\Delta y$, thus $N^{'}$~$Bin(n,1-F_{X_i}(y))$.
Thus, $f_{Y_{\alpha}}(y)=lim_{\Delta y ->0}\dfrac{(\binom{n}{\alpha -1}[F(y)]^{\alpha -1}[1-F(y)]^{n-\alpha})(F(y+\Delta y)-F(y))(\binom{n}{n-\alpha}[F(y)]^{\alpha}[1-F(y)]^{n-\alpha})}{\Delta y}$
Following this, I am not able to get the correct answer which is $\dfrac{n!}{(\alpha -1)!(n-\alpha)!}[F(y)]^{\alpha-1}[1-F(y)]^{n-\alpha}f(y)$
Can anyone help ?